Generally, for a prime $p$, Prüfer $p$ group is defined to be direct limit of the system of groups $\{\mathbb{Z}/p^n\mathbb{Z}\colon n\in \mathbb{N}\}$ with the homomorphisms $\mathbb{Z}/p^n\mathbb{Z} \rightarrow \mathbb{Z}/p^{n+1}\mathbb{Z}$ induced by multiplication by $p$. But on Wikipedia, it is defined with some "characterization":

It is the unique infinite $p$ group, in which every element has $p$ $p$'th roots.

It is the unique infinite $p$ group which is locally cyclic (every finite se of elements of group generates a cyclic group.)

How do we prove the "uniqueness" in these characterizations using the definition which is given in terms of direct limits?

I think you misunderstand something. To prove there is a unique blah, you prove that there is at least one blah (the direct limit for instance) and any two blah are isomorphic. The second part has nothing to do with the direct limit.
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Jack SchmidtJun 20 '11 at 13:59

@Jack: I don't think there is a substantial problem. One can prove that the direct limit construction satisfies the given property (your first clause), and then try to prove the uniqueness clause along the lines you suggest, but by appealing to the universal property of the direct limit somehow. I read the question as asking that.
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Arturo MagidinJun 20 '11 at 17:35

You can also define it to be the group of complex numbers which when raised to some $p$-power-th power yield 1.
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Scott CarnahanJun 20 '11 at 17:38

1

@Arturo, oh, like every group which satisfies it must be a direct limit, or something? Yeah, OK, that actually seems natural.
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Jack SchmidtJun 20 '11 at 17:38

1 Answer
1

First, some notation. Let us represent an element of $\mathbb{Z}/p^n\mathbb{Z}$ by the pair $(a,n)$, where $a\in\mathbb{Z}$; so $(a,n)=(b,n)$ if and only if $a\equiv b\pmod{p^n}$.

We write $\mathbb{Z}_{p^{\infty}}$ for the direct limit; the elements of the direct limit are equivalence classes of the form $[a,n]$, $a\in\mathbb{Z}$, $n\geq 1$, with $[a,n]=[b,m]$ if and only if (i) $(p^{m-n}a,m)=(b,m)$ if $m\geq n$; or (ii) $(a,n) = (p^{n-m}b,n)$ if $m\lt n$. Addition is defined by
$$[a,n]+[b,m] = \left\{\begin{array}{ll}
\ [a+p^{n-m}b,n] &\text{if }n\geq m;\\
\ [p^{m-n}a+b,m] &\text{if }n\lt m.
\end{array}\right.$$

The universal property of the direct limit means that if $G$ is any group, and we have group homomorphisms $f_n\colon \mathbb{Z}/p^n\mathbb{Z}\to G$ such that for all $n\leq m$, $f_n(a,n) = f_m(p^{m-n}a,m)$, then there exists a unique $f\colon\mathbb{Z}_{p^{\infty}}\to G$ such that $f[a,n] = f_n(a)$.

First, we need to show that $\mathbb{Z}_{p^{\infty}}$ satisfies the conditions given; then we want to show that given any group $G$ that satisfies the given conditions, we have an isomorphism from $G$ to $\mathbb{Z}_{p^{\infty}}$ (alternatively, show that $G$ has the universal property associated to the direct limit).

Property 1. The unique infinite $p$-group in which every element has exactly $p$ distinct $p$th roots.

Note that $\mathbb{Z}_{p^{\infty}}$ is infinite (it contains at least $p^n$ elements for each $n$. Also, each element has order a power of $p$, since $p^n[a,n] = [p^na,n] = [0,n]$ (since $p^na\equiv 0\pmod{p^n}$). So $\mathbb{Z}_{p^{\infty}}$ is an infinite $p$-group.

Also, every element of $\mathbb{Z}_{p^{\infty}}$ has at least $p$ distinct $p$th roots: let $[a,n]\in\mathbb{Z}_{p^{\infty}}$. Then for each $k$, $0\leq k\lt p$, we have
$$p[a+kp^n,n+1] = [pa+kp^{n+1},n+1] = [pa,n+1] = [a,n];$$
and $[a+kp^n,n+1] = [a+\ell p^n,n+1]$ if and only if $kp^n\equiv \ell p^n\pmod{p^{n+1}}$, if and only if $k\equiv \ell\pmod{p^n}$. Since we are assuming $0\leq k,\ell\lt p$, we conclude that $k=\ell$, so $[a,n]$ has at least $p$ distinct $p$th roots.

To show there are exactly $p$ distinct $p$th roots, first we show there are exactly $p$ distinct elements of order dividing $p$. If $p[b,m] = [0,m]$, then $pb\equiv 0\pmod{p^m}$, hence $p^{m-1}|b$. Writing $b=p^{m-1}k$ we have $[b,m] = [kp^{m-1},m] = [k,1]$. Thus, every element of order $p$ must be one of $[k,1]$ with $0\leq k\lt p$, so there are at most $p$ elements of order dividing $p$; each of these is distinct, so there are exactly $p$ elements of order dividing $p$.

Now suppose that $[b,m]$ is a $p$th root of $[a,m]$. Then for each $k$, $0\leq k\lt p$, we have
$$p\Bigl( [b,m] - [a+kp^n,n+1]\Bigr) = 0.$$
Therefore,
$$\Bigl\{ [b,m] - [a+kp^n,n+1]\Bigm| k=0,1,\ldots,p-1\Bigr\} = \Bigl\{ [0,1], [1,1],\ldots,[p-1,1]\Bigr\}$$
(since the $[b,m]-[a+kp^n,n+1]$ are pairwise distinct and all elements of order $p$). Therefore, there exists $k$ such that $[b,m]-[a+kp^n,n+1] = [0,1]$, hence $[b,m]=[a+kp^n,n+1]$, proving that the aforementioned $p$th roots of $[a,n]$ are the only $p$th roots of $[a,n]$. Thus, $[a,n]$ has exactly $p$ distinct $p$th roots. This proves that $\mathbb{Z}_{p^{\infty}}$ is an infinite $p$-group in which every element has exactly $p$ distinct $p$th roots.

The next step is to show that any infinite $p$-group in which every element has exactly $p$ distinct $p$th roots is isomorphic to the direct limit. Let $G$ be such a group.

First, we construct maps $f_m\colon \mathbb{Z}/p^m\mathbb{Z}\to G$ for each $m$ inductively, ensuring at each step that the maps form a consistent system. Since we do not know ahead of time whether $G$ is abelian, I will use multiplicative notation for $G$. For each $n\in\mathbb{N}$, let
$$G[p^n] = \{g\in G\mid g^{p^n}=1\}.$$
Note that $G[p]\subseteq G[p^2]\subseteq G[p^3]\subseteq\cdots$.

Assume that we have shown that $G[p^k]$ is a cyclic group of order $p^k$, generated by $x_k$ for $1\leq k\leq n$, and have defined maps $f_k\colon\mathbb{Z}/p^k\mathbb{Z} \to G[p^k]$ in such a way that the maps commute with the natural embeddings $\mathbb{Z}/p^k\mathbb{Z}\hookrightarrow \mathbb{Z}/p^{k+1}\mathbb{Z}$ for all $k$, $0\leq k\lt n$. In particular, we have that $x_{k+1}^p = x_k$ for each $k$.

Consider $G[p^{n+1}]$. Note that $g\in G[p^{n+1}]$ if and only if $g^p\in G[p^n]$. Since $G[p^n]$ has $p^n$ elements, and each element has exactly $p$ distinct $p$th roots, we conclude that $G[p^{n+1}]$ has $p\times p^n = p^{n+1}$ elements. Let $x_{n+1}$ be a $p$-th root of $x_n$. Then $x_{n+1}$ has order $p^{n+1}$ (since $x_n$ has order $p^n$), and since $\langle x_{n+1}\rangle \subseteq G[p^{n+1}]$, and both have $p^{n+1}$ elements, then $G[p^{n+1}]$ is a cyclic group, generated by $x_{n+1}$. Define $f_{n+1}\colon\mathbb{Z}/p^{n+1}\mathbb{Z}\to G[p^{n+1}]$ by mapping $[1,n+1]$ to $x_{n+1}$. This map is compatible with the embedding $\mathbb{Z}/p^n\mathbb{Z}\hookrightarrow \mathbb{Z}/p^n\mathbb{Z}$, and hence with all the previous embeddings as well. This completes the induction step.

Note that since $G$ is a $p$-group, if $g\in G$ then there exists $k$ such that $g\in G[p^k]$; thus, $G = \cup G[p^k]$.

Thus, we have a family of consistent maps $f_m\colon\mathbb{Z}/p^m\mathbb{Z}\to G$.

To show that $G$ has the universal property of the direct limit, let $H$ be any group, and let $h_m\colon\mathbb{Z}/p^m\mathbb{Z}\to H$ be a family of group homomorphisms which commute with the embeddings $\mathbb{Z}/p^k\mathbb{Z}\hookrightarrow \mathbb{Z}/p^{k+1}\mathbb{Z}$.

Note that $h$ is well-defined: if $g = x_k^r = x_m^s$, with $k\leq m$, then we know that $x_k = x_m^{p^{m-k}}$, hence $x_k^r = x_m^{rp^{m-k}}$. Hence $s\equiv rp^{m-k}\pmod{p^m}$ (since $x_m$ is of order $p^m$). Therefore,
$$h_k(r,k) = h_m(p^{m-k}r,m) = h_m(s,m)$$
which shows $h(g)$ is well defined (we have used that the $h_n$ are a consistent system, so that $h_k(r,k) = h_{k+1}(pr,k+1)$, etc).

Also, $h$ is a group homomorphism: if $g,g'\in G$, let $k$ be such that $g,g'\in G[p^k]$. Then $gg'\in G[p^k]$ (since $G[p^k]$ is a subgroup), and so
$$h(gg') = h_k(gg') = h_k(g)h_k(g') = h(g)h(g').$$
And for every $m$, we have $h_m(a,m) = h(f_m(a,m))$ by definition, so $h$ fits into the required commutative diagrams.

Now assume that $\psi\colon G\to H$ is any group homomorphism such that $h_m = \psi\circ f_m$ for each $m$. Then $\psi(x_k) = \psi(f_k(1,k)) = h_k(1,k) = h(x_k)$. Since $G$ is generated by the $x_k$, this proves that $\psi=h$.

Thus, $G$ has the properties of the direct limit of the system
$$\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p^2\mathbb{Z}\to\mathbb{Z}/p^3\mathbb{Z}\to\cdots$$
and therefore there exists a (unique) isomorphism $\mathbb{Z}_{p^{\infty}}\cong G$. Thus, $\mathbb{Z}_{p^{\infty}}$ is the unique (up to unique isomorphism) infinite $p$-group in which every element has exactly $p$ distinct $p$th roots, as claimed. QED

Property 2. The unique infinite $p$ group that is locally cyclic.

We already know that $\mathbb{Z}_{p^{\infty}}$ is an infinite $p$-group. To show it is locally cyclic, it suffices to show that any two elements are contained in a cyclic subgroup (since then the subgroup they generate is itself cyclic).

Now let $G$ be a locally cyclic infinite $p$-group. We will show that every element of $G$ has exactly $p$ distinct $p$th roots, which will establish the desired isomorphism using part 1.

Let $a\in G$, and let $|\langle a\rangle|=p^n$. Since $a$ is of finite order and $G$ is infinite, there exists $b\in G$, $b\notin \langle a\rangle$. In particular, the subgroup generated by $a$ and $b$ is cyclic, $\langle a,b\rangle = \langle c\rangle$, and $|\langle c\rangle| \gt p^n$. Thus, $\langle c\rangle\cong \mathbb{Z}/p^{n+k}\mathbb{Z}$ for some $k\gt 0$; in this subgoup $a$ has exactly $p$ distinct $p$th roots. Thus, every element of $G$ has at least $p$ distinct $p$th roots.

Now suppose that $r_1,\ldots,r_p$ are $p$ distinct $p$th roots of $a$, and let $s$ be any $p$th root of $a$. Then $\langle r_1,\ldots,r_p,s\rangle$ is cyclic of order a power of $p$, and contains $a$. Since in a cyclic group of order $p^n$ each element has at most $p$ distinct $p$th roots, then $s$ must be one of $r_1,\ldots,r_p$, proving that $a$ has exactly $p$ distinct $p$th roots.

Thus, each element of $G$ has exacty $p$ distinct $p$th roots, and so is isomorphic to $\mathbb{Z}_{p^{\infty}}$ by the first part. QED

Well, you'd only be accelerating the thermodynamical collapse of the universe if you did everything.
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Mariano Suárez-Alvarez♦Jun 21 '11 at 2:57

Slight strengthening: The Prufer group is the unique $p$-group in which every element has exactly $p$ distinct $p$th roots, i.e., such a group is forced to be infinite. Because taking $p$th powers is a function mapping the group to itself that is certainly not injective (due to multiple $p$th roots of any element). So if the group was finite, the function would not be surjective and there would be elements without $p$th root.
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PatrickRFeb 16 '13 at 22:40